Q#44
The door of an almirah is 6 ft high, 1.5 ft wide and weighs 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.Answer:
Let the forces exerted by each of the hinges on the door be F in horizontal direction and N in the vertical direction.
Equating the torques about the lower hinge,
F(4/3.28) = 8g(½)(1.5/3.28) {Since 1 m = 3.28 ft}
F = 9.8(1.5) = 14.7 N
Equating the forces in the vertical direction,
2N – 8g = 0
N = 4g
N = 39.2 N
Since F and N are mutually perpendicular, the magnitude of the resultant force by each of the hinges
= √(F² + N²) = √(14.7² + 39.2²)
= √1752.73 ≈ 42.0 N
Q#45
A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is θ.
Answer:
Let us draw a diagram showing forces on the rod. If m is mass of the rod then its weight mg acts at the middle point i.e. L/2 distance from the end. Let R be the normal reaction by the roller and NA by the ground. If µ be the coefficient of the friction between the rod and the ground then a force of friction, fA = µNA, acts at the lower end of the rod to check the movement.
Since the rod has no linear motion, hence the sum of forces in horizontal and vertical directions will be zero.
In the horizontal direction,
fA = NBsinθ
µNA = NBsinθ
NB = µNA/sinθ (i)
F(4/3.28) = 8g(½)(1.5/3.28) {Since 1 m = 3.28 ft}
F = 9.8(1.5) = 14.7 N
Equating the forces in the vertical direction,
2N – 8g = 0
N = 4g
N = 39.2 N
Since F and N are mutually perpendicular, the magnitude of the resultant force by each of the hinges
= √(F² + N²) = √(14.7² + 39.2²)
= √1752.73 ≈ 42.0 N
Q#45
A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is θ.
Answer:
Let us draw a diagram showing forces on the rod. If m is mass of the rod then its weight mg acts at the middle point i.e. L/2 distance from the end. Let R be the normal reaction by the roller and NA by the ground. If µ be the coefficient of the friction between the rod and the ground then a force of friction, fA = µNA, acts at the lower end of the rod to check the movement.
Since the rod has no linear motion, hence the sum of forces in horizontal and vertical directions will be zero.
In the horizontal direction,
fA = NBsinθ
µNA = NBsinθ
NB = µNA/sinθ (i)
In the vertical direction,
NB cosθ + NA = mg
µNAcosθ/sinθ + NA = mg
mg = NA(µcosθ + sinθ)/sinθ (ii)
Since the rod does not rotate, the sum of torques of all the forces will be zero. Hence taking moment about the lower end A, we get
NB (h/sinθ) – mg(L/2)cosθ = 0
NBh – ½mgLcosθsinθ = 0
2NBh = mgLcosθsinθ
2h(µNA/sinθ) = {NA(µcosθ + sinθ)/sinθ}(Lcosθsinθ)
[Putting the values of NB and mg from (i) and (ii)]
2hµ = (µcosθ+sinθ)(Lcosθsinθ)
2hµ = µLcos2θsinθ + Lcosθsin2θ
µ(2h – Lcos²θsinθ) = Lcosθsin²θ
µ = Lcosθsin²θ/(2h – Lcos²θsinθ)
Answer:
(a) Mass M = 300 g = 0.300 kg, Length L = 50 cm =0.50 m
The moment of inertia about the center I = ML²/12
The moment of inertia about an end I' = I + Mr²
{Where r = L/2 = 0.25 m}
= ML²/12 + Mr²
= 0.30(0.5)(0.5/12) + (0.30)(0.25)(0.25)
= 0.025 kg.m²
Angular speed = ⍵ = 2 rad/s
Hence the angular momentum of the rod about the axis of rotation
= I'⍵
= 0.025(2) = 0.05 kg-m²/s
(b) The speed of the center of the rod
v = ⍵r = 2(0.25) =0.50 m/s = 50 cm/s
(c) The kinetic energy of the rod
= ½I'⍵² = 0.50(0.025)(2)² J
= 0.50(4)(0.025) J
= 0.050 J
Q#47
A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.
Answer:
Mass M = 2.0 kg, Edge a = 10 cm = 0.10 m.
The moment of inertia about the diagonal = Ma²/12
= 2.0(0.10²)/12
= 0.01/6 =1/600 kg.m²
Torque = 0.10 N.m
Angular acceleration α = Torque/M.I.
= 0.10/(1/600) = 60 rad/s²
Angular speed at t = 5 s,
⍵ = 0 + 60(5) = 300 rad/s
Hence the angular momentum at this instant
= I⍵ = (1/600)(300)
= 0.50 kg.m²/s
Kinetic Energy
E = ½I⍵²
= ½(1/600)(300)²
= 75 J
Q#48
Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1.5 x 10⁸ km.
Answer:
Let the mass of the earth = M.
Its radius r = 6400 km = 64 x 10⁵ m, Radius of the orbit R = 1.5 x 10⁸ km = 1.5 x 10¹¹ m.
The angular speed of the spin of the earth
⍵ = (2π/24)3600 rad/s
The angular speed of the revolution of the earth
⍵' = (2π/365)(24)(3600) rad/s
If the moment of inertia of the earth about its axis = I = (2/5)Mr²
Then angular momentum = L= I⍵
The moment of inertia of the earth about the sun
I' = I + MR²
= (2/5)Mr² + MR² = M(2r²/5 + R²)
The angular momentum about the sun = I'⍵'
Hence the ratio of the angular momentum
= I⍵/I'⍵'
= [(2/5)Mr²⍵]/[M(2r²/5+R²)⍵']
= ⍵/(1 + 5R²/2r²)⍵'
= [2π/(24 x 3600)]/[(1 + 5(2.25 x 10²²)/(2 x 64 x 64 x 10¹⁰)(2π/365)(24)(3600)]
= 365/(0.00137 x 10¹²)
= 2.66 x 10⁻⁷
Q#49
Two particles of mass m₁ and m₂ are joined by a light rigid rod of length r. The system rotates at an angular speed ⍵ about an axis through the center of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is L = µr²⍵ where µ is the reduced mass of the system defined as µ = m₁m₂/(m₁ + m₂).
Answer:
Let the center of mass of the system from the mass m₁ be x, then x = m₂r/(m₁ + m₂)
The moment of inertia of the system about the CoM
I = m₁{m₂r/(m₁ + m₂)}² + m₂{m₁r/(m₁ + m₂)}²
= [m₁m₂² + m₁²m₂]r²/(m₁ + m₂)²
= [m₁m₂(m₁ + m₂)]r²/(m₁ + m₂)²
= m₁m₂r²/(m₁ + m₂)
= [m₁m₂/(m₁ + m₂)]r²
= µr²
Since the angular speed = ⍵
Hence the angular momentum Γ = I⍵ = µr²⍵.
Hence proved.
Q#50
A dumbbell consists of two identical small balls of mass ½ kg each connected to the two ends of a 50 cm long light rod. The dumbbell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5.0 N acts on one of the masses in the direction of its velocity for 0.10 s. Find the new angular velocity of the system.
Answer:
The moment of inertia of the system
I = 2mr²
= 2(½)(0.25)2
= 0.0625 kg.m²
angular speed of ⍵ = 10 rad/s
Time of impulse = t = 0.10 s
Torque of the impulsive force = T = 5(0.25) = 1.25 N.m
Impulsive torque = Tt
= 1.25(0.10)
= 0.125 Nm.s
change in angular momentum = Impulsive torque
If ⍵' is the new angular velocity of the system, then
I⍵' – I⍵ = Tt
⍵' = ⍵ + Tt/I
= 10 + 0.125/0.0625
= 12 rad/s
NB cosθ + NA = mg
µNAcosθ/sinθ + NA = mg
mg = NA(µcosθ + sinθ)/sinθ (ii)
Since the rod does not rotate, the sum of torques of all the forces will be zero. Hence taking moment about the lower end A, we get
NB (h/sinθ) – mg(L/2)cosθ = 0
NBh – ½mgLcosθsinθ = 0
2NBh = mgLcosθsinθ
2h(µNA/sinθ) = {NA(µcosθ + sinθ)/sinθ}(Lcosθsinθ)
[Putting the values of NB and mg from (i) and (ii)]
2hµ = (µcosθ+sinθ)(Lcosθsinθ)
2hµ = µLcos2θsinθ + Lcosθsin2θ
µ(2h – Lcos²θsinθ) = Lcosθsin²θ
µ = Lcosθsin²θ/(2h – Lcos²θsinθ)
Q#46
A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end, Calculate (a) the angular momentum of the rod about the axis of rotation. (b) the speed of the center of the rod and (c) it's kinetic energy. Answer:
(a) Mass M = 300 g = 0.300 kg, Length L = 50 cm =0.50 m
The moment of inertia about the center I = ML²/12
The moment of inertia about an end I' = I + Mr²
{Where r = L/2 = 0.25 m}
= ML²/12 + Mr²
= 0.30(0.5)(0.5/12) + (0.30)(0.25)(0.25)
= 0.025 kg.m²
Angular speed = ⍵ = 2 rad/s
Hence the angular momentum of the rod about the axis of rotation
= I'⍵
= 0.025(2) = 0.05 kg-m²/s
(b) The speed of the center of the rod
v = ⍵r = 2(0.25) =0.50 m/s = 50 cm/s
(c) The kinetic energy of the rod
= ½I'⍵² = 0.50(0.025)(2)² J
= 0.50(4)(0.025) J
= 0.050 J
Q#47
A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.
Answer:
Mass M = 2.0 kg, Edge a = 10 cm = 0.10 m.
The moment of inertia about the diagonal = Ma²/12
= 2.0(0.10²)/12
= 0.01/6 =1/600 kg.m²
Torque = 0.10 N.m
Angular acceleration α = Torque/M.I.
= 0.10/(1/600) = 60 rad/s²
Angular speed at t = 5 s,
⍵ = 0 + 60(5) = 300 rad/s
Hence the angular momentum at this instant
= I⍵ = (1/600)(300)
= 0.50 kg.m²/s
Kinetic Energy
E = ½I⍵²
= ½(1/600)(300)²
= 75 J
Q#48
Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1.5 x 10⁸ km.
Answer:
Let the mass of the earth = M.
Its radius r = 6400 km = 64 x 10⁵ m, Radius of the orbit R = 1.5 x 10⁸ km = 1.5 x 10¹¹ m.
The angular speed of the spin of the earth
⍵ = (2π/24)3600 rad/s
The angular speed of the revolution of the earth
⍵' = (2π/365)(24)(3600) rad/s
If the moment of inertia of the earth about its axis = I = (2/5)Mr²
Then angular momentum = L= I⍵
The moment of inertia of the earth about the sun
I' = I + MR²
= (2/5)Mr² + MR² = M(2r²/5 + R²)
The angular momentum about the sun = I'⍵'
Hence the ratio of the angular momentum
= I⍵/I'⍵'
= [(2/5)Mr²⍵]/[M(2r²/5+R²)⍵']
= ⍵/(1 + 5R²/2r²)⍵'
= [2π/(24 x 3600)]/[(1 + 5(2.25 x 10²²)/(2 x 64 x 64 x 10¹⁰)(2π/365)(24)(3600)]
= 365/(0.00137 x 10¹²)
= 2.66 x 10⁻⁷
Q#49
Two particles of mass m₁ and m₂ are joined by a light rigid rod of length r. The system rotates at an angular speed ⍵ about an axis through the center of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is L = µr²⍵ where µ is the reduced mass of the system defined as µ = m₁m₂/(m₁ + m₂).
Answer:
Let the center of mass of the system from the mass m₁ be x, then x = m₂r/(m₁ + m₂)
The moment of inertia of the system about the CoM
I = m₁{m₂r/(m₁ + m₂)}² + m₂{m₁r/(m₁ + m₂)}²
= [m₁m₂² + m₁²m₂]r²/(m₁ + m₂)²
= [m₁m₂(m₁ + m₂)]r²/(m₁ + m₂)²
= m₁m₂r²/(m₁ + m₂)
= [m₁m₂/(m₁ + m₂)]r²
= µr²
Since the angular speed = ⍵
Hence the angular momentum Γ = I⍵ = µr²⍵.
Hence proved.
Q#50
A dumbbell consists of two identical small balls of mass ½ kg each connected to the two ends of a 50 cm long light rod. The dumbbell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5.0 N acts on one of the masses in the direction of its velocity for 0.10 s. Find the new angular velocity of the system.
Answer:
The moment of inertia of the system
I = 2mr²
= 2(½)(0.25)2
= 0.0625 kg.m²
angular speed of ⍵ = 10 rad/s
Time of impulse = t = 0.10 s
Torque of the impulsive force = T = 5(0.25) = 1.25 N.m
Impulsive torque = Tt
= 1.25(0.10)
= 0.125 Nm.s
change in angular momentum = Impulsive torque
If ⍵' is the new angular velocity of the system, then
I⍵' – I⍵ = Tt
⍵' = ⍵ + Tt/I
= 10 + 0.125/0.0625
= 12 rad/s
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